Prove algebraically that the equation is an identity

Can someone please help me through this?

[csc(x)] / [cos(x)] - [cos(x)] / [sin(x)] = tan(x)

[csc(x)] / [cos(x)] - [cos(x)] / [sin(x)]

---csc(x) = 1 / [sin(x)], so [csc(x)] / [cos(x)] = 1 / [sin(x)cos(x)]

1 / [sin(x)cos(x)] - [cos(x)] / [sin(x)]

---multiply top and bottom of [cos(x)] / [sin(x)] by cos(x)

1 / [sin(x)cos(x)] - ([cos(x)]^2) / [sin(x)cos(x)]
(1 - [cos(x)]^2) / [sin(x)cos(x)]

---[sin(x)]^2 + [cos(x)]^2 = 1, so 1 - [cos(x)]^2 = [sin(x)]^2

[sin(x)]^2 / [sin(x)cos(x)]
[sin(x)] / [cos(x)]

--- tan(x) = [sin(x)] / [cos(x)]

tan(x)